Chapter 3: Electricity

Q
Electricity

Question:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

(i) For 9 Ω

We will connect two resistors $R_1$ and $R_2$  in parallel and the third resistor $R_3$ in the series with the parallel combination of $R_1$ and $R_2$.

The equivalent resistance of the parallel combination is given by,

$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_P} = \frac{1}{6} + \frac{1}{6}$

$\frac{1}{R_P} = \frac{1+1}{6}$

$\frac{1}{R_P} = \frac{2}{6}$

$\frac{1}{R_P} = \frac{1}{3}$

$R_P = 3\ \Omega$

Now, the equivalent resistance of the circuit,
$R_{eq} = R_P + R_3$

$R_{eq} = 3 + 6 = 9\ \Omega$

(ii) For 4 Ω

We will connect two resistors $R_1$ and $R_2$ in series combination and the third resistor $R_3$ in the parallel with the series combination of $R_1$ and $R_2$.

The equivalent resistance of the series combination is given by,

$R_S = R_1 + R_2$

$R_S = 6 + 6 = 12\ \Omega$

Now, the equivalent resistance of the circuit, $R_{eq}$ will be

$\frac{1}{R_{eq}} = \frac{1}{R_S} + \frac{1}{R_3}$

$\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6}$

$\frac{1}{R_{eq}} = \frac{1 + 2}{12}$

$\frac{1}{R_{eq}} = \frac{3}{12}$

$R_{eq} = \frac{12}{3}$

$R_{eq} = 4\ \Omega$

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